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TREES FOR /c-REVERSE MULTIPLES
Anne Ludington Young
Loyola College i n Maryland, B a l t i m o r e , MD 21210
(Submitted J u l y 1990)
Let x be an n - d i g i t , base g number
n- 1
(1)
X = ]Ta^
i= 0
with 0 < a^ < g and an-\
n- 1
(2)
kx = £
a
* 0.
If, for some integer k9 where 1 < k < g,
n-i-i9i
i= 0
then x is called a k-reverse
multiple,
Previously, this author showed that all
^-reverse multiples may be found using rooted trees [3]. A more detailed
examination of these trees is the focus of this paper.
If x is a ^-reverse multiple, then we obtain from (1) and (2) the following
equations
(3)
kai + ri_l
= an^l.
+ vig,
i
i = 0, . . ., n - 1,
where 0 < r^ < g for i = 0, ..., n - 2 and r_]_ = Pn-i
=
(3) gives aQ ^ 0 since #„-i * 0. To determine whether
multiples for a given g, we consider the equations in
the (i + l ) s t step, i = 0, 1, . .., we examine the pair
(4)
)
(fea„_1_i + Vyi-l-i
= a
i
+
r
0. Letting i = n - 1 in
there are any £:-reverse
(3) two at a time. At
of equations
'n-l-i9
seeking nonnegative integers a^, a n _i_^, r^ , and rn_2--z; less than g9 where
^i-\
and vn_i_^
are known from the previous step. The following graphical notation
an
is convenient. If P „ _ I _ ^ , ^ -1, a n _j_^, a^9 rn-2-i>
d ^ satisfy (4), then we
will write
O n - W
(5)
^i-l)
(a^-!-^ a,-)
(p M _ 2 _ i , r^)
(Implicit in this notation is the assumption that the a T s and r f s are nonnegative integers less than g.)
When a given g has fc-reverse multiples, we are able to generate a rooted
tree. We call the root of the tree O n - i , F-\) = (0, 0 ) , the 0 t h level and the
r
node designated by (rn_2-i:>
i)> t n e (^ + l ) s t level. Since 0 < ri < g > there
are only a finite number of possible distinct nodes. If a node is labeled with
a pair that has already appeared in the tree, the tree can be pruned. The
following theorem shows how a tree is used to determine ^-reverse multiples.
The proof appeared in [3] and hence is omitted here.
Theorem
1: For a given g, suppose there are A:-reverse multiples; that is, suppose a tree exists. There is a 2i + 2-digit or a 2i + 3-digit number satisfying (2) if and only if the tree contains at the i, i + 1, and i + 2 levels,
respectively,
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TREES FOR k-REVERSE MULTIPLES
On-l-i>
(sn-l-
n - l )
a
a
( n-l-i>
i)
"i-\>
(level i )
\®n - 1 - i» b-L)
(si t)
( r , r)
(level I + 1)
(B, B)
(t, s)
(level £ + 2)
where, in the second case,, B = {gs - t)/{k
t ) / (k -- 1).
r e s p e c t i v e l y , by
x = an^lan-2
a _i_,-aa -.
••• a
n - l - i 7i
.• .• .• aa\0-Q
^n
x = 2?n_12?n_2 . . . b
. . . bbxb
i
Yb^0
2 >n.nl.-iBb
l - - •Bbi
In these cases, x i s
given,
n = -2% + 2
n = 2i + 3.
D
Theorem 1 shows the connection between a rooted tree and /(-reverse multiples. A node of the form (i5, r) gives rise to a /(-reverse multiple with an
even number of digits. Consecutive nodes (s, t) and (t, s) produce a multiple
with an odd number of digits. The following example illustrates the use of
this theorem.
Example 1: g = 10, k = b.
We begin by letting vn-\
4a 0
+0
P_X = 0 in (4) and solve the system:
an_x + 1 0 P 0
an
+ 0.
The only solution is i)n-2=z 0? p o = ^ 5 a n - l = 2 5 and ag = 8. This gives the node
and edge labels for the first level of the tree. We continue in this manner
and obtain the following pruned tree:
(0,
0)
1 3)(2'
1 (1'
( 3 ,^ 3 )
8)
(0,
(6)
( 7 , 1)
(3,
7)
( 9 , 9)
0)
(3 , 3)
( 8 , 2)
(0
(0, 0)
(0,
>-
/
0)
0)
\ . ( 2 , 8)
( 0 , 3)
The tree is not continued any further since (0, 0 ) , (0, 3 ) , and (3, 3) have
appeared previously.
Observe that the node label (0, 0) follows (0, 0) at level 5, but not at
level 1. This will always be the case since the equations in (4) are satisfied
by the trivial or zero solution. Although P Q * 0, the node label (0, 0) is
permissible after the first level.
By Theorem 1, the node (3, 3) at the second level gives rise to the 4-digit
4-reverse multiple 2178. Moreover, the consecutive nodes (3, 3) and (3, 3)
produce the 5-digit multiple 21978. Extending this portion of the pruned tree
shows that all numbers of the form 219...978 are 4-reverse multiples. Thus,
there are n-digit 4-reverse multiples for all n > 4.
The relationship between the node and edge labels and verifying that a specific base g number x is, in fact, a /(-reverse multiple may be demonstrated by
performing base g multiplication of x by k, explicitly indicating all carries
from one digit to the next. It should be noted that when some x is known to be
a /(-reverse multiple this computation provides an alternate way to obtain some
of the node labels.
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167
TREES FOR k-REVERSE MULTIPLES
For example, 21782178 is a base 10 4-reverse multiple [corresponding to the
path from the root to node (0, 0) at level 4 in (6) above]. The multiplication verifying this fact is:
2
3
1
8
7
0
3
7
0
8
1 2
0
2
3
1
8
3
0
7
8
x 4
7
1 2
The carries from the node pairs and the digits of x form the edge labels.
From Theorem 1, the digits of 21782178 are the first elements of the edge
labels from the root to node (0, 0) at level 4 followed by the second elements
for the same edge labels taken from node (0, 0) at level 4 back up to the root.
Similarly, the carry numbers noted above the digits of x are the elements of
the node labels along the path. The first four carries are the first elements
of the node labels from level 1 to level 4, and the second four carries are the
second elements of the same node labels from level 3 back up to level 0. The
root is always labeled (0, 0) and by Theorem 1 (since 21782178 has an even
number of digits) the node label at level 4 must have both digits the same. D
The following examples illustrate some characteristics exhibited by trees
for ^-reverse multiples like the one shown in (6). We will use bold type for a
node that determines a fe-reverse multiple with an even number of digits and
underlining for one that determines a ^-multiple with an odd number of digits.
Further, since we recognize the existence of /^-reverse multiples graphically by
particular types of node labels, we will omit the edge labels. There is no
loss in doing this, for we may always use (4) to solve for
an. • l - i = (fer„-i -iff - krn-2--i H+ ri9 - Ti-OHM?- - 1),
at = (krig - kri - \ + r„_!-.i<3 - .-z-,)/^ 2 - 1).
le 2:
9 "•
1 1 , k = 7.
( 0 , 0)
( 1 , 5)
"(6,
(5,
( 0 , 0)
a, _5)
(5,
1)
6)
( 6 , Jl
By Theorem 1, the node (1, 5) along with its child (5, 1) determines a 3-digit
multiple and (6, 6 ) , a 4-digit one. Both (5, 1 ) , with its child (1, 5 ) , and
(6, 6 ) , with its child (6, 6 ) , give rise to 5-digit multiples. In fact, there
are n-digit 7-reverse multiples for n > 3. •
Example
3: g = 19, k = 14.
(0, 0)
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TREES FOR k-REVERSE MULTIPLES
In this case, there are n-digit 14-reverse multiples for n = 6 and n > 10. Q
Although we require r^ < g, in the examples above it happens that vi < k.
In [3] this was shown always to be the case.
In many instances the entire pruned tree can be determined from just an
initial branch. The following theorem gives one way in which this can be done.
Theorem
2:
s)
If
then
(v, u)
{b, a)
(a, b)
(u, v)
( s , r)
Proof: By hypothesis, the equations in (4) must be satisfied.
order of the two equations gives the desired result. Q
Switching the
As an illustration of Theorem 2, consider the tree in Example 3.
we know
(0, 0)
Suppose
I
(1, ID
(8, 13)
• I
(6, 6)
Then Theorem 2 allows us to derive
(6, 6)
(13, 8)
(11, 1)
I
(0, 0)
immediately without using (4).
We will use the notation
[r, s]
(7)
I [a, b]
[u, v]
to indicate solely that the equations in (4) are satisfied by integers; that
is,
(8).
Kb + s = a + vg,
ka + u = b + rg.
Thus, the notation in (7) does not imply that the integers are nonnegative and
less than g.
As before, when these latter restrictions do occur,, we will use
the (., .) notation instead of [., . ] .
The next two technical lemmas will be
useful in the theorems that follow.
Lemma
1:
Suppose there are integers such that
[r,
s]
[a, b]
[u, v]
with s, u < g and 0 < r, v.
Proof:
Then 0 < a, b.
Eliminating b from the equations in (8) and rearranging, we find
a(k2
- 1) = k(rg - u) + (vg -
Thus, given the hypotheses, 0 < a.
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s).
Similarly, 0 < b.
0
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TREES FOR fc-REVERSE MULTIPLES
2: Suppose there are integers such that
Lemma
[r,
s]
I [a, b]
[u,
v]
with
(0 < s, u,
p, v < ks
[? * k - l9 v * k - I, s * Q, or u * 0.
(10)
Then a, b < g.
From (9) we have
Proof:
a(k2
- 1) = krg - ku + vg - s
< g(kr + v)
< g(k(k - 1) + (fc - 1))
= g(kz - 1).
Given the restrictions in the third part of (10), one of the above two inequalities must be strict. Thus, a < g. Similarly, b < g.
•
3: If there are integers such that
Theorem
(P,
(11)
and
s)
(P',
I (a, b)
(u,
| (a', b>)
r
v)
vr)
(u ,
then
( p + v',
s +
(12)
sr)
sf)
(a + a ' , b + 2?')
r
(u + u ,
v + v ')
so long as
{
s + s', u + uT < g,
r + rr, v + vr < k,
r + rr * k - I, v + vf
* k - 1, s + s1 * 0, or u + ur * 0.
Proof: By hypothesis, (8) must be satisfied by P, s, . .., and by v', s f , ... .
Adding the corresponding equations gives the desired equations for (12). Since
all the numbers in (11) are nonnegative, those in (12) must be also. By Lemma
2, a + aT and b + br must be less than g. 0
Theorem 4: If there are integers such that
(P,
and
s)
(P',
I (a, 2>)
(13)
sr)
| (a', 6')
f
(w, u)
(u ,
v1)
then
(p - p ' , 5 - 5')
(14)
I {a - a\
(U
- Ur9
b -
bf)
Vr)
V -
so long as
fO < s - sr,
{0 < r - r'y
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TREES FOR k-REVERSE MULTIPLES
Proof: By hypothesis, (8) must be satisfied by p, s, . .., and by vf , sf, ... .
Subtracting the corresponding equations gives the desired equations for (14).
Since all the integers in (13) are less than g9 those in (14) must be also. By
Lemma 1, a - a' and b - b! are positive. •
The above theorems allow the completion of all or at least large portions
of a pruned tree when only an initial piece is known. Suppose, in Example 1,
only
(05 0)
(15)
(0, 3)
(3,' 3)
were known. We would be able immediately to derive the rest:
(3. 3)
(3, 0)
(3, 3)
(0, 0).
The left side follows from Theorem 2; the right from Theorem 3, since
(0, 0) and
(3, 3)
(0, 3)
(3, 0)
imply
(3, 3)
(3, 3).
Note that by the restrictions in Lemma 2,
(0, 0) and
(3, 0) do not imply
1
(0, 3)
(0, 0)
(3, 0)
(0, 3).
Thus, we are able to derive the entire pruned tree for Example 1 knowing only
(15) or5 equivalently, knowing only that 2178 is a 4-reverse multiple. Similarly , a careful examination of the trees in Examples 2 and 3 shows that each
follows5 respectively, from the 3-digit number 118 and the 6-digit number
1 2 11 8 17 15.
It may sound very restrictive to assume that we know an initial portion of
a tree. However, this is equivalent to assuming that a /c-reverse multiple for
a given g is known. The problem then is to find or characterize all other
multiples and this is done using the associated pruned tree. Hence, if we know
an n-digit fe-reverse multiple for some small n, then we do know an initial
portion, of the tree. The problem is then to complete the tree quickly and
easily. As an illustration, consider the following, more complicated, example.
Example 4: g = 44, k = 27.
The 6-digit number
(16)
1 7 18 5 24 31
is a base 44, 27-reverse multiple; this can be verified through multiplication:
1
31
11
7
24
3
18
5
15
5
18
19
24
31
x 27
7
1
The carry numbers are numbers in the node labels of the initial portion of the
tree, so we have
1992]
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TREES FOR ^-REVERSE MULTIPLES
(0, 0)
(4, 19)
. I
(11, 15)
(3, 3)
We complete the tree using the above theorems.
For example, by Theorem 3
(4, 19) and (3, 3) imply (7, 22)
I
I
I
(11, 15)
(15, 11)
(26, 26)
In the tree that follows, the superscript j on a particular node indicates that
it was derived using Theorem J, J = 2, 3, 4. So in the above case, we would
write (26, 2 6 ) 3 .
(0, 0)
I
(4, 19)
(11, 15)
(17)
(4, 1 9 ) 3 (11, 1 5 ) 2
Note that the theorems above do not guarantee that the pruned tree of (17)
is complete and that no branches are missing. The next theorem addresses this
concern.
Theorem
tains
(18)
Suppose g has a /(-reverse multiple; further, suppose the tree con-
5:
(?>
(a, b) y / \
s)
(u, v)
where u > u!
(a', b')
r
(u ,
and V < V'.
v!)
Then
[0, 0]
[a - a', b - bf]
[u - ur, v -
vr]
where 0 < u - ur < k, -k < v - v ' < 0, -g < a - ar < 0 and -g < b - b! < 0.
Proof: Recall that if
must be less than k
Thus, all the claims
b - b! < 0. From (9)
a - a1 = (-k(u
r
Hence, a - a
< 0.
(18) occurs in a tree, then each number in the node label
and each number in the edge label must be less than g.
in the conclusion follow immediately except for a - ar,
we know that
- ur)
v')g)/(k2
+ (v r
Similarly, b - b
< 0.
1).
•
Suppose, for a given g, that we know some ^-reverse multiple and thus are
able to obtain the initial portion of the tree. We apply Theorems 2, 3, and 4
whenever possible until all branches end with nodes that have appeared previously. At this point, we are in the position of asking if there are any
missing branches. By Theorem 5, if there are no integers c, d, t , w for which
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TREES FOR fc-REVERSE MULTIPLES
[0, 0]
(19a)
| [-e, -d]
[t, -w]
where
(19b)
0 < t < k,
0 < w < k,
0 < c < g,
0 < d < g,
then we can be assured that there are no missing branches in the tree.
In all the examples considered thus far, (19) is never satisfied.
ify this for Example 1, we must consider the equations
(20)
-kd
-he
To ver-
= -c - lOw
+ t = -d.
obtained from equations (4). Eliminating d in (20) gives 4t = 15c - lOw; thus,
5|t. However, 0 < t < 5. Consequently, there are no solutions to (20) and,
hence, to (19). Thus, by Theorem 5, the tree in (6) is complete.
Theorem
6:
Suppose g has a ^-reverse multiple and the tree contains
(r, s)
(a, b)
(u,
v)
Further, suppose
[0, 0]
[-o,
[t,
-d]
-w]
with 0 < t < k,
0 < w < k,
(p, s)
or
(a - c,
(u + t ,
0 < c < g,
b - d)
v - w)
and 0 < d < g.
Then
(r, s)
\ (a + c,
(u - t ,
b + d)
v + w)
so long as either the three conditions u + t < g9 0 < V - w, and 0 < r or the
two conditions 0 < u - t and V + w < k are fulfilled.
Proof:
The first piece follows from Lemma 1; the second from Lemma 2. Q
The following example illustrates the use of Theorem 6.
Example
5: g = 40, k = 13.
The 5-digit number
(21)
2 24 30 1 34
is a 13-reverse multiple. As in Example 4, the number in (21) gives the initial portion of the tree which has node labels (0, 0 ) , (8, 11), and (9, 0) .
There is just one solution to (19); namely,
[0, 0]
-10]
We now use Theorems 2, 3, 4, and 6 to complete the tree:
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173
TREES FOR ^-REVERSE MULTIPLES
The double bar edges leading to nodes without a superscript indicate that none
of the above theorems apply. In these cases the nodes were found using (4) .
Note that there are only 3 such instances. On the other hand, the 16 superscripted nodes were found easily using the theorems indicated by the superscript as in Example 4.
As we have noted, there is just one solution to (19). We used this solution in conjunction with Theorem 6 to find 3 nodes. If the tree contained any
missing nodes, then by Theorem 5 equations (19) would have another solution.
Since, that is not the case, the tree is complete.
Acknowledgment
The author is grateful to the referee for many helpful suggestions.
References
1.
2.
3.
4.
174
C. A. Grimm & D. W. Ballew. "Reversible Multiples." J. Rec. Math.
8(19751976):89-91.
L. F. Klosinski & D. C. Smolarski. "On the Reversing of Digits." Math.
Mag.
42 (1969):208-10.
Anne Ludington Young. "/c-Reverse Multiples." Fibonacci
Quarterly
30.2
(1992):126-32.
Alan Sutcliffe. "integers That Are Multiplied When Their Digits Are Reversed ." Math. Mag. 39 (1966):282-87.
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